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Where is the incenter?

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  • John Conway
    I was trying to prove for myself the assertion that the locus of the incenter over all triangles with given Euler segment is the interior of the
    Message 1 of 3 , Jun 1, 2000
      I was trying to prove for myself the assertion that the locus
      of the incenter over all triangles with given Euler segment is
      the interior of the orthocentroidal circle, or "arena". However,
      the algebra turns out to be a bit too tough even for me - I can
      prove that the incenter is always inside this circle, but not that
      all points of it are reached.

      So I hope that one of the hyacinths (Paul perhaps?) can help by
      finding the expression for the squared distance from the incenter
      to the midpoint of GH as a function of a,b,c.

      I'm also interested in the locus of the excenters. I thought
      at first that they would all lie outside the arena, but seem to
      have disproved this. There are two problems:

      1) what's the set of points where any excenter can be?

      2) supposing a > b > c, what are the individual loci of
      the a-, b-, and c- excenters?

      I suspect that the answer to 1) has a fairly simple shape,
      while those that arise in 2) might be quite complicated.

      Regards to all, John Conway
    • Lambrou Michael
      ... It seems vectors can do the jog without too much trouble (but I confess I did not do the dirty work to the last line): square of IM = inner product of (
      Message 2 of 3 , Jun 2, 2000
        >
        > So I hope that one of the hyacinths (Paul perhaps?) can help by
        > finding the expression for the squared distance from the incenter
        > to the midpoint of GH as a function of a,b,c.
        >
        It seems vectors can do the jog without too much trouble (but I confess I
        did not do the dirty work to the last line):
        square of IM = inner product of ( (g+h)/2-i) by itself,
        where g=(a+b+c)/3 etc all known. Save work by taking a=0 (where of course
        I have changed notation to vectors, easily going back to lengths)

        Michael
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